Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
H2(x, C2(y, z)) -> H2(C2(S1(y), x), z)
H2(C2(S1(x), C2(S1(0), y)), z) -> H2(y, C2(S1(0), C2(x, z)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
H2(x, C2(y, z)) -> H2(C2(S1(y), x), z)
H2(C2(S1(x), C2(S1(0), y)), z) -> H2(y, C2(S1(0), C2(x, z)))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
H12(C2(S1(x), C2(S1(0), y)), z) -> H12(y, C2(S1(0), C2(x, z)))
H12(x, C2(y, z)) -> H12(C2(S1(y), x), z)
The TRS R consists of the following rules:
H2(x, C2(y, z)) -> H2(C2(S1(y), x), z)
H2(C2(S1(x), C2(S1(0), y)), z) -> H2(y, C2(S1(0), C2(x, z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
H12(C2(S1(x), C2(S1(0), y)), z) -> H12(y, C2(S1(0), C2(x, z)))
H12(x, C2(y, z)) -> H12(C2(S1(y), x), z)
The TRS R consists of the following rules:
H2(x, C2(y, z)) -> H2(C2(S1(y), x), z)
H2(C2(S1(x), C2(S1(0), y)), z) -> H2(y, C2(S1(0), C2(x, z)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.